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Albinochka1901
@Albinochka1901
July 2022
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Помогите решить неравенства: 2cos(pi-x)<=1 и -4tg(x+pi/8)<=1
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nKrynka
1) 2cos(π - x) ≤ 1
cos(π - x) ≤ 1/2
cos(x - π) ≤ 1/2
arccos(1/2) + 2πn ≤ x - π ≤ 2π - arccos(1/2) + 2πn, n∈Z
π/3 + 2πn ≤ x - π ≤ 2π - π/3 + 2πn, n∈Z
π/3 + 2πn ≤ x - π ≤ 5π/3 + 2πn, n∈Z
π/3 + π + 2πn ≤ x ≤ 5π/3 + π + 2πn, n∈Z
4π/3 + 2πn ≤ x ≤ 8π/3 + 2πn, n∈Z
2) - 4tg(x + π/8) ≤ 1
tg(x + π/8) ≥ - 1/4
arctg(-1/4) + πn ≤ x ≤ π/2 + πn, n∈Z
- arctg(1/4) - π/8+ πn ≤ x ≤ π/2 - π/8+ πn, n∈Z
- arctg(1/4) - π/8 + πn ≤ x ≤ 3π/8 + πn, n∈Z
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Answers & Comments
cos(π - x) ≤ 1/2
cos(x - π) ≤ 1/2
arccos(1/2) + 2πn ≤ x - π ≤ 2π - arccos(1/2) + 2πn, n∈Z
π/3 + 2πn ≤ x - π ≤ 2π - π/3 + 2πn, n∈Z
π/3 + 2πn ≤ x - π ≤ 5π/3 + 2πn, n∈Z
π/3 + π + 2πn ≤ x ≤ 5π/3 + π + 2πn, n∈Z
4π/3 + 2πn ≤ x ≤ 8π/3 + 2πn, n∈Z
2) - 4tg(x + π/8) ≤ 1
tg(x + π/8) ≥ - 1/4
arctg(-1/4) + πn ≤ x ≤ π/2 + πn, n∈Z
- arctg(1/4) - π/8+ πn ≤ x ≤ π/2 - π/8+ πn, n∈Z
- arctg(1/4) - π/8 + πn ≤ x ≤ 3π/8 + πn, n∈Z